A lot of people are asking if you can get the intrinsic metric tensor matrix without referencing the "outside space" (XYZ). We can get the intrinsic metric from several places: 1. using the extrinsic basis 2. invent one from our imagination 3. obtain it from another equation, like the Einstein Field Equations in physics Once we have the intrinsic metric, we can do ALL calculations involving distances and angles completely intrinsically. But we need the intrinsic metric first. Sometimes this involves getting some help from the outside space at the start.

@whdaffer1

2 ай бұрын

Couldn't you use special coordinates with the radius r=constant? So you would only have phi=longitude and phi=co-latitude. 2d and completely intrinsic.

I have to thank you - I'm a Physics Engineer and I was never shown how tensors work. I have been looking for course notes online, videos and the like, but this is the first time I can truly say I understand them. Keep it up, Chris.

Thanks again chris for your work! You are building up the material really beatifully! I'm always looking foreward to the next video :-)

The art of teaching! Dissecting the body of Tensors with surgical precision. Charismatic teacher.Thank you for making such a difficult subject understandable to us.

@eigenchris

5 жыл бұрын

Thanks. :) I might disagree with the "charismatic" part. But I will deliver monotone whenever you like.

@user-vm9zt6tm3h

5 жыл бұрын

I suppose there is a jacobian matrix from (X,Y,Z) space to (u,v) space of 3*2 dimension and due to Lecture11 at 13.22 min we can find the metric tensor g of (u,v) space byJ(transpose)*g(X,Y,Z)J . Obviously like the existance of a metric tensor field in (u,v) space there must be a Jacobian field in the same space.

@eigenchris

5 жыл бұрын

Yes, you are correct.

These videos are such an invaluable resource. Step by step, straight to the point, but still thorough enough to avoid confusion. I applaud you!

Thank you for sharing the course link! This and your video series are helping me a lot!

Wow! That video was really an eye opener for me. Thank you so, so much for this series!

After the surprise of discovering that the partial derivatives are the same as basis vectors, this lecture is an absolute gem. and to me fascinating. When you were finished with the Jacobian matrix, you definitely turned the intimidating Jmatrix into a simple tool. I looked at your references to the math 430 course and found this material to be heavy-handed when compared against your bite-sized crystal clear coverage of each topic. I am taking copious notes and screen shots so it will be some time yet before I finally catch up with your last lecture but I am definitely enjoying this journey.

Some really good graphics here. I hope people can appreciate how much work you've put into making these videos.

Thank you "eigenchris" for amazing video lectures about tensors. recommending and using your lectures for teaching. I hope there is no issue of sharing your lectures for teaching purposes.

Really excellent work, enjoying these lectures, fun to watch.

ThankU for taking from Your truely valuable time to post and share.

I really love these presentations, thanks for your effort.

GREAT EXPLANATION AND GRAPHICS....LOOKING FORWARD TO THE FOLLOW-UP VIDEOS TO THIS VIDEO....

Really great job at explaining this difficult subject. Thank you!

Finest quality work. Great job!

This has really inspired me to delve all the way in. I've started working through An Introduction to Manifolds and Differential Geometry both by Loring Tu.

@astrolillo

2 жыл бұрын

Good luck in your studies, never give up

Thanks for the series, it would be super helpful if you can suggest accompanying exercises (from books or anywhere) to go along with the videos.

Dear Chris This one was a beautiful class

@eigenchris

5 жыл бұрын

Thanks. I was excited to make this one.

I'd love to see you teaching differential geometry! Things like the geometric intuition of covariant derivative, geodesics, tangent bundles and so on. :) btw,: best videos in tensors and tensor calculus on youtube!

@eigenchris

5 жыл бұрын

Thanks. Vidoes on covariant derivative and geodesics are coming soon. Probably September. I'm already making slides.

@keyyyla

5 жыл бұрын

eigenchris fantastic!

@frankdimeglio8216

2 жыл бұрын

@@eigenchris UNDERSTANDING WHAT IS THE COMPARATIVE AND BALANCED BODILY/VISUAL EXPERIENCE OF THE MAN WHO IS STANDING ON WHAT IS THE MOON: The BULK DENSITY of the Moon is comparable to that of (volcanic) basaltic LAVAS on the Earth. The energy density of LAVA IS about three times that of water. SO, now, get a good and CLEAR look at what is the ORANGE SUN. The human body is, in fact, about as dense as water ! Look up at the blue sky. Importantly, this is a balanced BODILY/VISUAL EXPERIENCE !! The Moon is ALSO BLUE, AS it is barely visible then; AND the Earth is ALSO BLUE. SO, let's now match up (on balance) the eye, the blue sky, AND the ORANGE "Sun"/LAVA. Notice the basic match in sizes and brightness when considering the SETTING AND fully illuminated Moon AS WELL. (Most of the human body is water, with an average of roughly 60 percent. The average density of the Moon is also 60 percent in comparison with that of the Earth.) Beautiful. Relatively speaking, the terrestrial moon is, in fact, roughly a match (regarding it's size, at 27 percent) WITH the surface land area of the Earth (at 29 percent). In conclusion, in establishing the comparative AND BALANCED BODILY/VISUAL EXPERIENCE of gravity on what is the Moon, we would multiply one HALF times one third in order to get the correct answer. This explains quantum gravity, AS E=MC2 IS F=ma; AS ELECTROMAGNETISM/energy is gravity; AS gravity/acceleration involves BALANCED inertia/INERTIAL RESISTANCE. Accordingly, the rotation of WHAT IS THE MOON matches it's revolution !!! Great. Consider the relation with the tides. The Moon is ALSO BLUE. E=MC2 IS F=ma ON BALANCE. This NECESSARILY represents, INVOLVES, AND DESCRIBES what is possible/potential AND actual IN BALANCE, AS ELECTROMAGNETISM/energy is gravity. Gravity IS ELECTROMAGNETISM/energy. In conclusion, I have mathematically proven that the gravity of what is THE MOON is ONE SIXTH in direct comparison with what is THE EARTH/ground; AS gravity/acceleration involves BALANCED inertia/INERTIAL RESISTANCE; AS E=MC2 IS F=ma; AS GRAVITATIONAL force/energy IS proportional to (or BALANCED with/as) inertia/INERTIAL RESISTANCE. Gravity IS ELECTROMAGNETISM/energy. Think QUANTUM GRAVITY !!! Think energy DENSITY. It all CLEARLY makes perfect sense ON BALANCE. The ENERGY DENSITY of LAVA is about THREE TIMES greater than water. There would be no gravity on the blue Earth, AS the EYE then stands in balanced relation WITH what is the BLUE sky !! BALANCED BODILY/VISUAL EXPERIENCE does then dictate that the gravity of WHAT IS THE MOON would be reduced by (and at) one half, thereby bringing the surface gravity of what is THE MOON to BALANCE with that of THE EARTH/ground at one sixth. So, one half (ON BALANCE) of one third is one sixth. The orange "Sun" AND what is THE MOON ARE BALANCED with what is THE EYE. (Think about what is LAVA.) Notice, comparatively, that these forms or manifestations ARE semi-detached in relation to touch/tactical EXPERIENCE !! Moreover, the Moon is TERRESTRIAL in COMPARISON; SO (ON BALANCE) the Moon is ALSO then about one quarter (at 27 percent) the size of EARTH/ground !! The blue Pacific OCEAN takes up about HALF of the view of what is the Earth ON BALANCE !! SO, we WOULD multiply ONE HALF (ON BALANCE) in order to account for WHAT IS THE MOON, this ORANGE "Sun", AND LAVA on balance !! Now, in conclusion, the land surface area of the Earth is 29 percent. This is exactly ON BALANCE WITH BOTH one third AND one fourth, AS ELECTROMAGNETISM/energy is gravity ! E=MC2 IS CLEARLY F=ma ON BALANCE ! Gravity IS ELECTROMAGNETISM/energy. Great. By Frank DiMeglio

Thank you so much! It is really an incredibly clear lecture!

Muchísimas gracias profesor por sus videos!!!

@supramayro434

Жыл бұрын

Sus 😳

Many thanks for your excellent work.

sir -- u have given a lot of information -- lot of information is difficult to retain-- please make short lectures -- other wise u r excellent --thank u sir

it is convenient to represent the sphere as a parametrized circle in the xz plane (sin (v), 0, cos (v)) rotated in the xy plane by matrix with parameter u

Briliant!! Thank you

Very good

Laying it down! Thanks - really enjoyed this. Ahh for me: derivative operators as vectors.

Thanks again for this :D

Great video :)

Wow! @2:05 it took me a while, but i understand the full transformation! its important to have the depiction of the Z,Y,X axis superimposed on the sphere, but the dashed circle got in the way of my thinking because i thought it might be specific to the y-axis (put that dashed circle in the southern hemisphere!) do not crowd your pictures diagrams! Thank you, i think i grew a few iq points from this vid alone!

Excellent!!!

Awesome. Thank you

22:32 In study of parametric surface R(u,v) and computation of 1st fundamental form, we always work like this. It seems to me that it is a mixture of intrinsic and extrinsic approach.

Amazing!

Excellent video, all the playlist so far is awesome, do you know some place i can find exercises to pratice the contents of this video?

@eigenchris

2 жыл бұрын

Sorry, I don't. One exercise to try is to try and derive the metrics for the cylinder and the saddle that I show briefly. For the cylinder, use the equations (θ,z) -> (x=cosθ, y=sinθ, z=z). For the saddle, use (x,y)-> (x,y,z=xy). You can try computing the tangent vectors and then the dot products of the tangent vectors to get the metrics.

What about treating the sphere as a 2 dimensional smooth manifold (well, actually as a riemaniann manifold, since we're talking about metric tensors)? I've always thought that this is the intrinsic way of looking at such geometrical objects.

@eigenchris

5 жыл бұрын

You are right. I'm trying to not complicate things too much right now. When it comes to manifolds, there are a lot of heavy definitions that are needed to discuss them formally (charts, topological spaces, homeomorphisms). I think all of this is sort of "overkill" when you are just starting out. I am trying to get to the core of the idea of intrinsic geometry without all the heavy definitions. The basic idea is that you have a space where the metric tensor changes from point to point. That's all I wanted to convey right now.

Awesome videos... While waiting for more, I'm re-watching them... I like the bit of coding magic with your lambda symbol at 1:37. Would you consider sharing the code?

@eigenchris

5 жыл бұрын

It's not my code. You can do it with any file at www.maptoglobe.com.

@eigenchris To perform the intrinsec method, you needed the transformation rule from the 2D plane to the 3D object. Without that given, is it possible to find out which 3D object we live on by doing a cartography of it, leading to the metric tensor field of that object? Do you know what shape does the scientific community think of the universe/space-time field? I love your videos! They make me thinking a lot about other questions.

@eigenchris

5 жыл бұрын

Yes, it is possible to know you are in a curved space without leaving the space. You can often figure it out by drawing triangles. On a flat space, the 3 angles of a triangle will add yp to 180 degress. On a sphere, they will add up to more than 180. On a saddle, they will add up to less than 180. As for how to get the metric tensor 100% intrinsically, I am not sure how to go about this... usually I see the metric tensor defined FIRST, and you just need to accept that it describes the space properly. In the case of General Relativity, you use Einstein's equations to solve for the metric tensor. The amount lf energy and momentum in the area determine thr metric tensor, and so determine how much the space is curved. As for thr overall shape of the universe, I have no idea. I recently saw a Minute Physics video that said the universe, overall, is flat, but with curved areas in small regions due to energy/momentum in the local area.

@swalscha

5 жыл бұрын

@eigenchris This method of triangles seems pretty useful indeed. I personally can't think of the universe as flat, at least as an ellipsoïd squished along one axes. Thanks for the response.

Is there any use cases of doing tensor calculus in non-commutative rings? Instead of having a strictly symmetric metric tensor, would there be any use of generalizing it to non-symmetric tensors as a result of non-commutativity?

@eigenchris

5 жыл бұрын

I have no idea. The standard formulation of general relativity uses a symmetric metric tensor. There's a wikipedia article call "Nonsymmetrical gravitiational theory" that indicates theories which involve a non-symmetric metric tensor but I haven't looked into that at all.

5:40 holy mother of math that's alot of equations !

Are we allowed put the indices on the row vector d/du_i instead of d/du^i to indicate it's a row vector instead of a column vector? I know it isn't super important but I feel like it's easier to distinguish. What do you think?

@eigenchris

5 ай бұрын

An upper index in the denominator behaves like a lower index. Hopefully that's satisfying enough of an answer.

Hey these videos are great! But I have a question: you say that the formula you get for the magnitude of dR/dlambda squared is only in terms of u, v and t, so you conclude it’s an intrinsic way of calculating distances, but you calculated the metric tensor by taking partials of R, which is defined in the dimensions higher than the (u,v) space. Doesn’t this make it not completely intrinsic? Or am I missing something? Thanks in advance to anyone for the help!

@eigenchris

3 жыл бұрын

For intrinsic spaces, you either have to be given the metric from the start, or get it from a formula like the Einstein Field Equations (where mass/energy/momentum determine the metric). Once you have the metric, you know pretty much every intrinsic fact about the surface. In this video we use the outside space to get the intrinsic metric, but all calculations afterward, like curvature, can be done completely intrinsically.

This is good but most manifolds are defined implicitly. One needs to know how to derive the metric tensor for implicit surfaces. From z=f(x,y) you can parametrize as (x,y,z(x,y)) and compute the metric tensor from that.

At 21:00 you say that in intrinsic view we can't reference anything outside of our curved space. But just earlier you derived intrinsic basis vectors by using embedded 3D space outside the sphere

@eigenchris

4 жыл бұрын

We need to get a way of measuring distances somehow. In some cases we start with extrinsic space, get the metric, and then forget about the extrinsic space. In other cases we just make up a metric for the intrinsic space and never use extrinsic space at all.

@DioD3

4 жыл бұрын

@@eigenchris Thank you.

11:19 though you mentioned in the description that dR/du should in fact be ∂R/∂u (similar for v), all three X, Y and Z are also multivariable functions of u,v (i.e. X(u,v), Y(u,v) and Z(u,v)), hence throughout the video ∂X/∂u, ∂X/∂v should be used instead of dX/du, dX/dv and so on.

jesus! you made some major lightbulbs go off in my dim mind!

At 9:46, shouldn’t the expansion of R with respect to (u,v) coordinates contain partial derivatives of R with respect to (u,v) ( not total) in the right hand side of the concerned equation?

@eigenchris

4 жыл бұрын

Yes, that was a mistake.

Nice video, but why are all these calculations irrelevant to the sphere radius? Why only for radius=1?

Now I'm wondering a thing: Given a metric tensor in u,v, how would you tell if the (u,v)-plane is flat or curved? And if it is curved, how would you tell if it can be embedded in a 3D space?

@eigenchris

5 жыл бұрын

To tell if a space is curved, you can compute something called "Gaussian Curvature", (usually denoted by "K"). This is a number that can tell you if the surface is positively curved (like a sphere), negatively curved (like a saddle), or not curved (like a flat plane, or a cylinder--a cylinder is considered "not curved" by mathematicians since it can be unfolded into a flat plane without any geometric distortion). The "Theorema Egregium" says that the Gaussian curvature is an intrinsic quantity that doesn't depend on the surrounding space. All this is slightly beyond my current knowledge so you'll have to google more. I would strongly consider checking out all 4 pdfs I show at the beginning of the video if you want to learn more details. As for knowing whether or not it can be embedded in a 3D space, I'm not sure. I think the "real projective plane" is an example of a 2D curved space which cannot be embedded in 3D. I believe the Klein Bottle is the same way. But I'm not sure how to prove that.

@maigowang

5 жыл бұрын

eigenchris Thanks so much for the pointers!

Moreover the object is oblivious of the fact that it is living in embedded surface.So the object shouldn't be describe by two parameteric equations???

Can you please suggest any tensor book? I want to study tensors but a very good book.

@eigenchris

5 жыл бұрын

As I mentioned in the video, there are some course notes on the metric tensor here: liavas.net/courses/math430/ Unfortunately I'm not aware of any textbooks on tensors.

@gurejalectures

5 жыл бұрын

eigenchris ok. Your method 9f teaching is very good. GOD bless you. Which software do you use to make your videos?

@eigenchris

5 жыл бұрын

I'm glad you find these helpful. I make slides in Microsoft Powerpoint (which has a built-in equation editor and pen tool). I hit the record button, and then export to video once I'm done.

@mythic3187

5 жыл бұрын

I'm reading student guide to vectors and tensors by Daniel fleisch.. also watching alex flournoys lectures on GR. I highly recommend..btw, thanks Eigenchris

@nilsfollmann2170

5 жыл бұрын

This is a great video series on tensor alculus. he has also written a book that goes with the videos, which fills the gap between multivariable calculus and most tensor books or the treatment of tensors in GR books. kzread.info/dash/bejne/l2SZrLqtpLO4iqw.html

Hate to be that guy but thought you might want to have it pointed out: 9:47 should have partial dR/du and dR/dv. Also, great video. This is the one I’ve been waiting for. I have to ask though, how would one go about finding the tangent space basis without assuming the existence of a higher dimension?

@eigenchris

5 жыл бұрын

You're right about the partial derivatives. It's a minor enough error that I don't think I'm going to bother fixing it. I just added a note in the description. Finding the tangent basis can always be done if you have coordinates defined in a space. (i.e. I can take the partial derivatives of a function along the U coordinate working purely in 2D space, 3D space is not required.) Unless you're asking about how to compute the dot product/metric tensor components? I'm not 100% sure how to answer this. I know in General Relativity, computing the metric tensor is the goal of Einstein's equations. You start with the known energy and momentum in your space, and then compute the metric tensor from that. Then, using the metric tensor, you can compute the lengths of curves in the space. I'm not sure how the metric is computed in pure math cases without reference to a 3D space. Sometimes the metric is just defined out of nowhere, and people start studying the geometry that the metric creates. Other times the metric is inspired by a known 3D shape. You might be interested in reading about non-euclidean geometry and look at some of the metric examples from that. One example: en.wikipedia.org/wiki/Poincaré_disk_model#Metric

@Cosmalano

5 жыл бұрын

eigenchris that is what I meant and I’ll definitely take a look at that, thanks. As far as I knew however we used the christoffel symbol to define intrinsic geometries, and as they are composed of the derivatives of the metric tensor itself, and they themselves characterize the Ricci tensor field, I had assumed you used the energy and momentum tensor to define the local curvature, and then derive the metric tensor out of the geodesic equation/christoffel symbols. Is this not correct?

@eigenchris

5 жыл бұрын

I'm honestly not that confident with my knowledge of General Relativity, or Christoffel Symbols. I've scene videos where the instructor takes the Schwarzchild metric from GR and derives geodesics from it. This is indeed done with the geodesic equations, which contain Christoffel Symbols, which as you say come from the derivatives of the metric. I have not yet gone through a derivation of the Schwarzchild metric so I'm not familiar with how it is obtained. I do plan on doing at least 2-3 videos on Christoffel Symbols and the Covariant derivative in August, though. I'll be filling in my knowledge as I go.

@Cosmalano

5 жыл бұрын

eigenchris I assume you saying “not confident” is just you being humble. When I studied all this stuff years ago it just never clicked until I watched your tensor algebra video series, which in my mind makes you one of the most capable teachers of the subject. Thanks to you I was finally able to read through Gravitation this summer (still have a long way to go, but it’s entirely coherent at least now). So thank you. I kept this comment in mind and looked for the best help I could find. This video is the second in a three part series on a derivation of the Schwarzschild metric. In it he takes the Christoffel symbols and uses them to calculate the Ricci tensor, based on substituted values for the components of the metric tensor that he then solves for themselves using the Ricci tensor. It helped me understand how to work with Christoffel symbols better so I’m hoping it helps you just as much/gives you the confidence boost you need, plus it’s the derivation of the Schwarzschild metric you were looking for (I hope). kzread.info/dash/bejne/fYOXkpuoldGces4.html

@drlangattx3dotnet

4 жыл бұрын

I have relly enjoyed these lectures. I have been jumping around with other teachers. Frederick Schuller does a good job too. The TpM basis can be found intrinsically. See Lecture 5 by Frederick Schuller. kzread.info/dash/bejne/opmkt8Fwfq25d7A.html

But if we do Intrinsic geometry, in order to find Metric Tensor field, we still need to do the forward transformation from metric tensor of 3d cartesian coordinate, which means we still need to calculate transformation matrix. Which is only possible if we know the relation between 3D coordinates and 2D coordinate.

@eigenchris

3 жыл бұрын

For intrinsic geometry, you either need to be given the metric already, or you need a way of getting the metric through an equation, like the Einstein aField Equations in General Relativity.

@sanidhyasinha5735

3 жыл бұрын

@@eigenchris oh okay, thanks

Can all intrinsically definend 2d surfaces be mapped to their 3d analog?

@eigenchris

Жыл бұрын

I don't think so. First exception that comes to my mind is the 2D hyperbolic plane. We have several "models" of it (several 2d coordinate systems with different metrics for getting the lengths and angles correct), but no 3D surface corresponds to the entire 2D hyperbolic plane.

At 22'40'' I am not sure the problem is solved by getting rid of R. You still have X, Y, and Z and d/dX, d/dY and d/dZ. These are a basis for a 3-dimensional tangent space.

How far away is this from the metric tensor used in general relativity.

@eigenchris

4 жыл бұрын

The metric tensor in GR is 4x4 rather than 2x2, since spacetime is 4-dimensional. Also, the GR metric tensor measures spacetime intervals, which can be negative ("timelike"), positive ("spacelike") or zero ("lightlike"). This is different than ordinary distances in 2D or 3D space, which are always positive. Other than that, the math behind them is basically the same. The metric tensor is a matrix of dot products of the basis vectors and helps measure distances along curves.

i need examples whit numbers ,thank you

Is there no way of retrieving the intrinsic metric tensor without using extrinsic geometry?

@eigenchris

4 жыл бұрын

You need to get the intrinsic metric tensor somehow. Either you take one inspired from extrinsic geometric, or you need to make it up some other way. In relativity, the metric tensor comes from the solution of Einsteins Field equations, which depend on the energy and momentum in the surrounding spacetime.

@kostasch5686

4 жыл бұрын

@@eigenchris Oh wow, 2 minute response. Thanks! Btw you re doing a great job with the series. I m binging it at the moment and enjoying it far more than a netflix show. Please know your work is acknowledged!

can we use spherical coordinate system to find out length of curve??? then how we will find derivative of r, thita and phi w.r.t lambda?

@eigenchris

4 жыл бұрын

You could. You'd just need to use the metric tensor for spherical coordinates. That is a 3x3 diagonal matrix whre the entries are 1, r^2 and r^2 (sin theta)^2.

hello plz tell me. when are you going to upload next videos.??

@eigenchris

5 жыл бұрын

Hopefully in the next few days.

The metric tensor itself is invariant right? What you mean by metric tensor field then the ‘components’ of the metric tensor itself is a field where in every point in the fied, the value can be different. So basically metric tensor field tells us about what the component of the metric tensor is in a given location.

@eigenchris

4 жыл бұрын

A metric tensor field gives you a metric tensor at every point, the same way a vector field gives you a vector at every point. The metric tensor at a point p is invariant, but its components change in different coordinate systems.

@RizkyMaulanaNugraha

4 жыл бұрын

eigenchris thanks for your very quick reply. I didn’t expect I will get a prompt comment :D. I’m sorry, I think my statement above is a little bit ambiguous. So I want to reclarify again. Do any of my following statements below false/are incorrect: 1. metric tensor are locally invariant. Only the components changes under basis changes, but the tensor is the same. 2. metric tensor fields ‘can’ have different tensors in different location, but all of them are locally invariant under basis changes. 3. tranformation rules between chosen basis A to basis B are globally invariant. It’s expressions are going to be the same. All tensors in this tensors fields will have the same transformations rules if we are still transforming between basis A to basis B (coordinate system A to coordinate system B). 4. even though transformation rules are globally invariant, it’s transformation matrice/tensor are only locally invariant. (As it is shown in the polar coordinate’s metric tensor, which comes directly from applying its Jacobian matrice, twice) Again thanks for the videos. I was surprised that the next videos played instantly, which means you don’t put ads. Do you have patreon account?

18:34 message to all Flat Earthers!!

22:30 Conceptually, this is such a small but crucial point. The last couple of days I've been breaking my head over the idea that d/dx and d/dy are defined as the basis for tangent vectors. I see how the operator obeys vector axioms, but, why an operator anyway? It's such a bizarre construct that looks at odds with elementary examples such a velocity. It's just an abstraction that removes the extrinsicity; when we get to apply the theory to a physical problem, the vector quantity (in this case R) will turn up (if I got everything right).

@eigenchris

3 жыл бұрын

Have you watched this earlier video? I try to justify why use treat derivative operators as vectors: kzread.info/dash/bejne/iHyfrpeBoK-WeLw.html

@orktv4673

3 жыл бұрын

@@eigenchris Going through it right now, and things are coming together again. This is convincing that velocity is indeed a tangent vector created by parametrizing position (we've just kept the position vector itself out of the picture; it's not essential to the calculus). I am now inclined to ask how this construction works for a vector field like the electric field, although I suspect that that is in fact a one-form (since it is obtained by the gradient i.e. exterior derivative of the potential). If that is correct, is there any other quantity that I should see as a tangent vector?

@eigenchris

3 жыл бұрын

@@orktv4673 I think just about any vector quantity can be viewed as a one-form if you apply the metric tensor to it (and every one-form can be viewed as a vector if you apply the inverse metric tensor). Each point on the manifold has its own tangent space where vectors of any type can live. Each point also has its own "cotangent space" where one-forms of any type can live. The electric field is maybe not the best example because in relativity it is not often interpreted as a vector, but instead as a "bivector" called the "Faraday Tensor" or "Electromagnetic Field Tensor". But again, at a given point, this will live in the tangent spaces. I hope this isn't too confusing for you. You might want to read up on it more.

@orktv4673

3 жыл бұрын

@@eigenchris Yes, vectors and covectors can be easily converted by applying the metric tensor, but the thing I'm looking at is the fact that the velocity vector can be explicitly constructed by taking the directional derivative of position. My point with the electric field was that taking directional derivatives is obviously amounts to the gradient. When Carroll discusses tangent vectors he vaguely "identifies" tangent spaces with derivative operators (as if the derivatives are merely an effective analogy, rather than the consequence of an explicit construction). Anyway, thanks for your videos mate.

The way that you transformed the uv plane into the sphere using X Y Z coordinates, can the same be done for transforming the uv plane into some X Y coordinates (Where X(u,v) and Y(u,v)) such that this space is physically distorted. Is there a way to find the basis vectors and find the metric? Thanks

@eigenchris

3 жыл бұрын

You can always change coordinates however you like. You can get the basis vectors in the new coordinate system by using partial derivativ3s with respect to the new coordinate variables. The metric tensor components can obtained from the dot products of these basis vectors.

@joshuapasa4229

3 жыл бұрын

@@eigenchris I might have phrased it wrong, but what I ment was a physical transformation that distorts th u v plane such that after the transformation the point (u,v) has moved somewhere else (X(u,v),Y(u,v)). A bit like how a 3d transformation physically distorts the u,v plane. Thanks

@joshuapasa4229

3 жыл бұрын

@@eigenchris Is it the same method as you have shown in your video?

@eigenchris

3 жыл бұрын

@@joshuapasa4229 Sorry, I thought I replied to this. Yes, you can always change the u,v-plane using the same way I have shown in the video.

@joshuapasa4229

3 жыл бұрын

@@eigenchris Ok thank you!

metric ，the ruler for the space

These videos are true treasure. One thing is bugging me though. When treating derivative operators as basis vectors you "loose" their length which is one of the properties of basis vectors. So being said that this is something mathematicians do to make things more rigor, it only means that I am missing something basic. Can you help me on this?

@eigenchris

3 жыл бұрын

I think you can still define a dot product (metric tensor) operation between partial derivative operators and get their "length". In differential geometry you often and up getting the lengths of curves by doing an integral over infinitesimally small length segments. You can use the metric tensor to get these infinitesimal lengths and integrate over them.

@djordjekojicic

3 жыл бұрын

@@eigenchris thanks for the quick answer which makes perfect sense because differentials are infinitesimal differences and "infinitesimals or infinitesimal numbers are quantities that are closer to zero than any standard real number, but are not zero" as one definition says. But sometimes I still find it hard to wrap my head around it.

Sir why did you use total differential in case of intrinsic for dR/d(lembda). you used (dR/du)(du/d(lembda)). use use d instead of partial derivatives for dr/d(lembda)?

@eigenchris

Жыл бұрын

There are likely some typos. For taking the derivative along a 1D path (single variable), I should use "d". For taking the derivative with respect to a coordinate variable in 2D space like u or v, I should use a "partial" derivative.

@NeerajVerma786

11 ай бұрын

@@eigenchris Thanks for making these videos. The way you teach maths like not just derivations but you also tells the geometrical meaning behind them is awesome not many people teach maths in this way. Thanks again a lot...

Hey, are you Canadian? I just noticed "zed".

@davidreichert9392

3 жыл бұрын

Maybe I only noticed it being Canafian myself, but his overall accent is very Canadian. I'm guessing from Ontario or the Maritimes.

At 1:01, can’t see how a cylinder (described with 3 coordinates rho, phi, z), a sphere (r,theta,phi) and a hyperbolic paraboloid (visibly shown in a 3-D cuboid) are 2-D surfaces. Perhaps I’m missing something conceptually...

@eigenchris

4 жыл бұрын

They are 2D because a bug living on them only has two main directions to walk in: the forward-backward direction and the left-right direction. The surfaces live in 3D space, but they can be parameterized in terms of 2 coordinates only (u and v, as shown with the sphere).

@anamikashukla1810

4 жыл бұрын

Thanks

Can you clarify the use of dR/du and partial derivatives in regards to your error? I see that both are used at times.

@eigenchris

4 жыл бұрын

They should all be partial derivatives, as the vector depends on both U and V.

@drlangattx3dotnet

4 жыл бұрын

@@eigenchris at 5:56 there are dX and dY and dZ used. These are correct? Can you explain a little bit here ? Thanks for videos.

@eigenchris

4 жыл бұрын

dX/d lambda is fine, because X is a function of a single variable (lambda). We use a partial derivative when a function depends on mumtiple variables.

@drlangattx3dotnet

4 жыл бұрын

@@eigenchris So EVERYWHERE dR /d lamda should be changed to partials? Sorry to be dense.

can one use this formulas in hyperbolic spaces? If so, how?

@eigenchris

3 жыл бұрын

The formulas are the exact same. The thing that differentiates a Euclidean space from a hyperbolic space is the metric tensor itself. So given the metric tensor, you just use the formula and you don't need to worry about what type of space you're in.

Why nobody mentioned 9:30

You said eu dot ev is same as ev dot eu, does it mean the metric tensor is a symmetric matrix always?

@eigenchris

3 жыл бұрын

Yes.

Can we somehow obtain the intrinsic metric tensor without using the standard basis?

@eigenchris

3 жыл бұрын

We can get the intrinsic metric from several places: 1. using the extrinsic basis 2. invent one from our imagination 3. obtain it from another equation, like the Einstein Field Equations in physics Once we have the intrinsic metric, we can do ALL calculations involving distances and angles completely intrinsically. But we need the intrinsic metric first. Sometimes this involves getting some help from the outside space at the start.

@charlie2720

3 жыл бұрын

@@eigenchris Aight, got it! Thank you!

I think the limits of the integral(s) are not really correct. missing 2*pi*r?

@eigenchris

3 жыл бұрын

Can you give a timestamp for the part you are talking about?

Is it possible to compute the dot product of the intrinsic basis vectors without making use of the x,y,z basis vectors?

@eigenchris

4 жыл бұрын

In pure mathematics, you either have to use the "extrinsic" space (xyz), or simply make up what the intrinsic metric tensor is on your own. In the case of general relativity, the metric tensor is determined by the amount of energy and momentum in the local region of spacetime.

@jackjones5592

4 жыл бұрын

But doesn‘t that mean eventhough I‘m a small person living on the 2D surface of the sphere and want to know everthing about that surface „intrinsicly“, I would still have to know the extrinsic 3D parametrisation of the sphere in order to compute the intrinsic metric tensor?

@eigenchris

4 жыл бұрын

@@jackjones5592 I've been thinking about this question for a while. You should be able to figure everything out If you're a tiny person living in a curved space with a bunch of coordinate lines drawn on the ground. These coordinate lines would have little "ticks" on them telling you were values of 1.0, 2.0, 3.0, 3.5, 3.72 are and so on. You would need to carry around some kind of measuring sticking around with you, where the measuring stick has length 1. Then, after picking some point to stand on, and assuming the small region around you is mostly-flat, you could take derivatives of a position vector with respect to the coordinate lines, you could get a set of basis vectors at that point. You could then measure how long those basis vectors are using your special unit measuring stick. These measurements would give you the diagonal elements of the metric tensor (e_i dot e_i). You could also use a protractor to measure the angles between each pair of basis vectors. Using this, and the basis vector lengths, you could compute the dot products between any pair of basis vectors, which are the off-diagonal elements of the metric tensor (e_i dot e_j). Is this a satisfying answer?

@jackjones5592

4 жыл бұрын

I think I understand your method of measuring the components, but then I would only have the metric tensor in the specific point I chose, wouldn‘t I? So I would have to measure every single point in order to get the full tensor field? Actually I posted this question before I was watching your intrinsic covariant derivative video, where you explicitly stated that it is not possble to compute the metric tensor(field) of a curved intrinsic space and you would need something to tell you what the components are like energy and momentum does in the case of GR. Still bothers me though that there is no other way of computing the dot products of some basis vectors. Wonder how this is handled outside of GR. Anyway I really like your series, I‘ve been looking for something like this for a while. Keep up the good work man! :)

Something doesn't seem quite right. How can the length on a spherical surface (1.12389) be less than the length on a flat surface (1.4142)? Think about it this way: if I travel on a flat surface and that flat surface is then deformed into a hill, will I really travel less going over the hill than when going on a flat surface?

@eigenchris

3 жыл бұрын

The u2 coordinate doesn't represent physical length; it's only the angle of longitude. The length it corresponds to increases at the equator amd decreases near the poles, so each "square" on the sphere will have a different area, depending on the distance from the equator.

@joeboxter3635

3 жыл бұрын

MYSTERY SOLVED.

How does one get the transformation rules for x y and z?

@eigenchris

4 жыл бұрын

What do you mean by transformation rules?

@lourencoentrudo

4 жыл бұрын

@@eigenchris I mean the equations by which u,v gets mapped out onto the spherical surface

Uh... in this example isn’t the intrinsic metric tensor only in terms of only the intrinsic dimensional variables in the same function because of how u=v..... it makes no difference what is plugged in computationally..

9:42 I don’t understand how can we expand a tangent vector using flat space coordinates? The flat space on the left is just flat, ie. a 2D sheet. If we we will use basis vectors of this flat space to expand any vector, we will just land on that space again, without noticing any curvature. I kind of follow the reasoning after that point, but I don’t understand why expanding the tangent vector in UV (or even calling this expansion a “tangent vector”), is a good idea. A 2D vector can’t be tangent to a flat 2D space, can it? What am I missing? PS. And thanks again for all your work on this!

@imaginingPhysics

2 жыл бұрын

The 2D map with this metric (derived from the 3D sphere) is not flat. And the tangent vectors of this 2D map-world do not live "on the map itself", techically they live in a "plane of their own" as differential operators. One might visualize them using a 3D space but the point is that it is not necessary. This is a bit tricky concept. Hope this helps.

@przadka

2 жыл бұрын

@@imaginingPhysics thanks for reminding me about me comment/question from 10 months ago :) i appreciate you took time to explain this - it is clear for me now. for some reason, when watching this video, i kept thinking that UV space is flat and of course it is not. it is just a R^2 representation of a surface of s sphere and and such it will carry the sphere's curvature.

should'n the metric tensor for sphere 3-by-3 matrix?

@eigenchris

5 жыл бұрын

In the "external" 3D space, we can use a 3x3 matrix for the metric tensor (it would be the identity matrix in cartesian coordinates). If we're looking at the "intrinsic" 2D surface of the sphere, we use a 2x2 matrix.

@sufyannaeem2121

5 жыл бұрын

@@eigenchris Is that mean christoffel symbols will also be different for both case you mentioned???

@eigenchris

5 жыл бұрын

@@sufyannaeem2121 Yes. In the external 3D cartesian space, the Christoffel symbols are all 0. In the intrinsic 2D space with the coordinate system on the sphere, the Christoffel symbols are non-zero.

I don't understand how the u and the v in the (intrinsic, 2D) u-v plane match up with the u and v in the (extrinsic, 3D) spherical representation. In the u-v plane, they are distances (or, numbers that represent distances), and in the sphere, the u is the *angle* from the north pole, and v is the *angle* from a point on the equator that is (X,Y) = (1,0). Someone care to clarify? Thanks!

@eigenchris

3 жыл бұрын

You can think of the intrinsic UV plane as being like a 2D "map" of the sphere's surface. All the information about the surrounding 3D space has been removed.

@destructionman1

3 жыл бұрын

@@eigenchris Thanks for responding! Hmm ok so, in the u-v intrinsic plane, are u and v distances or angles?

@eigenchris

3 жыл бұрын

@@destructionman1 They're neither. In more abstract spaces, coordinates don't need to have a physical interpretation, thry just mark points in the space with unique numbers.

Intrinsic vs extrinsic

Is Cylinder really a curved surface ? I would say no, as the metric tensor is constant everywhere.

@eigenchris

4 жыл бұрын

You're correct. It was easy to graph though.

I dont fully understand why u and v in 2d surface become angle in x,yz plane (3 d), I kow cos 0 is 1 but what relation with curve in 3d or ball? and you said x = cosvsinu why not cosusinu or sinvsinu? how you formed it

@eigenchris

4 жыл бұрын

I'm sorry but I don't understand your question. I explain the parametric equations for the sphere around 2:30. Is something unclear?

@zzzoldik8749

4 жыл бұрын

@@eigenchris u and v in 2D become cos (v)sin(u) in 3D. Become the angle (v) or (u)

Is there a book on this?

@eigenchris

4 жыл бұрын

You'd want to find a book on "differential geometry". I can't recommend anything specific. I learned mostly through course notes notes online.

geodesics for when, please????

@eigenchris

5 жыл бұрын

They are coming. Probably early September.

while using intrinsic approach you used the same parametric equations, you used for extrinsic approach.Also you used R which is an extrinsic approach.Why???

@eigenchris

4 жыл бұрын

In the intrinsic approach, I am able to express everything in terms of u amd v. I meantion near the end of the video that we should not be using the R vector if we want to be completely intrinsic... this is why the basis vector becomes d/dx instead of dR/dx.

@sufyannaeem2436

4 жыл бұрын

@@eigenchris but what about parametric equations???

@eigenchris

4 жыл бұрын

What do you mean?

@sufyannaeem2436

4 жыл бұрын

means you used three parametric equations for intrinsic approach??? I didn't understand why??

@eigenchris

4 жыл бұрын

​@@sufyannaeem2436We use the extrinsic space to get the metric tensor, but after that, we use U and V coordinates only. The metric tensor has to come from somewhere, but once we have it all the geometry is intrinsic.

Great, Thanks a lot, dear, Is there any Email address to share my idea with you. Regards

9:24 he's so cute

I didn’t know that there was calc 12

This is my version of Parametric Equation of a Sphere. ↓ X = cos(ν)sin(μ) Y = sin(ν)sin(μ) Z = 1(ν)cos(μ) Beautiful. Don't you think?

@eigenchris

5 жыл бұрын

That works. :)

Any brothers-flatearthers here?

@ES-qe1nh

Жыл бұрын

Lol